What is #f(x) = int 1/(x+3)-1/(x-4) dx# if #f(-2)=3 #? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Trevor Ryan. Jan 9, 2016 #therefore f(x)=ln|x+3|-ln|x-4|+3-ln6#. Explanation: #f(x)=int(1/(x+3)-1/(x-4))dx=int1/(x+3)dx-int1/(x-4)dx# #=ln|x+3|-ln|x-4|+C#. But since #f(-2)=3# is given as an initial/boundary condition, we can substitute it into the integral to find the constant of integration : #therefore ln|-2+3|+ln|-2-4|+C=3# #therefore0+ln6+C=3,=>C=3-ln6#. #therefore f(x)=ln|x+3|-ln|x-4|+3-ln6#. Answer link Related questions How do you find the constant of integration for #intf'(x)dx# if #f(2)=1#? What is a line integral? What is #f(x) = int x^3-x# if #f(2)=4 #? What is #f(x) = int x^2+x-3# if #f(2)=3 #? What is #f(x) = int xe^x# if #f(2)=3 #? What is #f(x) = int x - 3 # if #f(2)=3 #? What is #f(x) = int x^2 - 3x # if #f(2)=1 #? What is #f(x) = int 1/x # if #f(2)=1 #? What is #f(x) = int 1/(x+3) # if #f(2)=1 #? What is #f(x) = int 1/(x^2+3) # if #f(2)=1 #? See all questions in Evaluating the Constant of Integration Impact of this question 1199 views around the world You can reuse this answer Creative Commons License